Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 10005, M = 30;
using pii = pair<int, int>;
int n, bs[N], bnum[N], x[N], y[N], flx[N], fly[N], b[N];
vector<pii> ans, opt;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x[i];
  for (int i = 1; i <= n; i++) cin >> y[i];
  memset(bnum, -1, sizeof(bnum));
  for (int i = 1; i <= n; i++)
    for (int j = M, cur = 0; j >= 0; --j)
      if (x[i] & (1 << j)) {
        if (bs[j])
          x[i] ^= x[bs[j]], cur ^= flx[j];
        else {
          bnum[i] = j;
          bs[j] = i;
          flx[j] = cur | (1 << j);
          break;
        }
      }
  int idx = 0;
  for (int i = 1; i <= n; i++) {
    int fl = 0;
    for (int j = M; j >= 0; --j)
      if (y[i] & (1 << j)) {
        if (bs[j])
          y[i] ^= x[bs[j]], fl ^= flx[j];
        else {
          puts("-1");
          return 0;
        }
      }
    if (bnum[i] != -1) {
      b[idx] = i;
      fly[idx++] = fl;
      continue;
    }
    ans.push_back({i, i});
    for (int j = M; j >= 0; --j)
      if (fl & (1 << j)) ans.push_back({i, bs[j]});
  }
  for (int i = 0; i < idx; i++) {
    int tmp = fly[i];
    fly[i] = 0;
    for (int j = 0; j < idx; j++)
      if (tmp & (1 << bnum[b[j]])) fly[i] |= 1 << j;
  }
  for (int i = 0; i < idx; i++) {
    if (!(fly[i] & (1 << i)))
      for (int j = i + 1; j < idx; j++)
        if (fly[j] & (1 << i)) {
          opt.push_back({b[i], b[j]});
          opt.push_back({b[j], b[i]});
          opt.push_back({b[i], b[j]});
          swap(fly[i], fly[j]);
          break;
        }
    if (fly[i] & (1 << i)) {
      for (int j = i + 1; j < idx; j++)
        if (fly[j] & (1 << i)) opt.push_back({b[j], b[i]}), fly[j] ^= fly[i];
    }
  }
  for (int i = 0; i < idx; i++) {
    if (!(fly[i] & (1 << i))) ans.push_back({b[i], b[i]});
    for (int j = i + 1; j < idx; j++)
      if (fly[i] & (1 << j)) ans.push_back({b[i], b[j]});
  }
  reverse(opt.begin(), opt.end());
  for (auto &i : opt) ans.push_back(i);
  cout << ans.size() << endl;
  for (auto &i : ans) cout << i.first << ' ' << i.second << endl;
  return 0;
}